Chain rule of partial derivatives!

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Homework Statement


Suppose f(x,y) = 2x^5 + 4xy + 2y^3
g1(u,v) = u^2 - v^2
g2(u,v) = uv
h(u,v) = f(g1(u,v), g2(u,v))

Use chain rule to calculate:
dh/du (1,-1) and dh/dv (1,-1)


Homework Equations





The Attempt at a Solution



i let h (u,v) = 2(u^2 - v^2)^5 + 4(u^2-v^2)(uv) + 2(u-v)^3
then i tried drawing a chain rule tree diagram:

h --> f --> g1 and g2

g1 --> x and y

g2 --> x and y

so dh/du = (dh/df . df/dx . dx/du) + (dh/df . df/dy . dy/du)

then i got stuck :S

help please?
 

Answers and Replies

  • #2
vela
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Stuck how?
 
  • #3
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im not sure if my chain rule is correct to begin..i found this in my textbook though on calculating partial derivatives with changes in variables:

dh/du = dh/dg1 . dg1/du + dh/dg2 . dg2/du

so i followed that instead...
and got this

since h (u,v) = 2(u^2 - v^2)^5 + 4(u^2-v^2)(uv) + 2(u-v)^3

dh/dg1 = 10(u^2-v^2)^4 + 4uv

dg1/du = v

is that right so far?
 
  • #4
vela
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h is just a different name for f(g1(u,v), g2(u,v)), so the factor of dh/df in your original attempt is unnecessary.

Since you wrote out h(u,v) explicitly in terms of u and v, you can simply differentiate it as you normally would do to find the answer you should get.

What's probably confusing you are all the variables. So try it like this:
[tex]h = f(g_1, g_2) = 2g_1^5 + 4g_1g_2 +2g_2^3[/tex]so[tex]\frac{\partial h}{\partial g_1} = 10g_1^4 + 4g_2[/tex]and then express the g's in terms of u and v. (This gives you what you got above.) Do the same thing with g2. Then put those together with the partial derivatives of the g's to form the final result. (You got [itex]\partial g_1/\partial u[/itex] incorrect.)
 

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